Descriptional Complexity of Formal Systems (Draft) Deadline for submissions: April 14, 2009 Final versions: June 15, 2009

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1 DCFS 2009 Descriptional Complexity of Formal Systems (Draft) Deadline for submissions: April 14, 2009 Final versions: June 15, 2009 Quotient Complexity of Regular Languages Janusz Brzozowski David R. Cheriton School of Computer Science University of Waterloo Waterloo, ON Canada N2L 3G1 Abstract. The past research on the state complexity of operations on regular languages is examined, and a new approach based on an old method (derivatives of regular expressions) is presented. Since state complexity is a property of a language, it is appropriate to define it in formallanguage terms as the number of distinct quotients of the language, and to call it quotient complexity. The problem of finding the quotient complexity of a language f(k, L) is considered, where K and L are regular languages and f is a regular operation, for example, union or concatenation. Since quotients can be represented by derivatives, one can find a formula for the typical quotient of f(k, L) in terms of the quotients of K and L. To obtain an upper bound on the number of quotients of f(k, L) all one has to do is count how many such quotients are possible, and this makes automaton constructions unnecessary. The advantages of this point of view are illustrated by many examples. Moreover, new general observations are presented to help in the estimation of the upper bounds on quotient complexity of regular operations. Keywords: automaton, derivative, expression, language, operation, quotient, rational, regular, state complexity, upper bound 1 Introduction It is assumed that the reader is familiar with the basic concepts of regular languages and finite automata, as described in many textbooks. General background material can be found in Dominique Perrin s [25] (1990) and Sheng Yu s [30] (1997) handbook articles; the latter has an introduction to state complexity. A more detailed treatment of state complexity can be found in Sheng Yu s survey [31]. The present paper concentrates on the complexity of basic operations on regular languages. Other aspects of complexity of regular languages and finite automata are discussed in [2, 6, 9, 14, 16, 18, 28, 29]; this list is not exhaustive, but it should give the reader a good idea of the scope of the work on this topic. This research was supported by the Natural Sciences and Engineering Research Council of Canada under grant no. OGP

2 2 J. Brzozowski 2 State complexity or quotient complexity? The English term state complexity of a regular language seems to have been introduced by Birget 1 [1] in 1991, and is now in common use. It is defined as the number of states in the minimal deterministic finite automaton (DFA) accepting the language [31]. There had been much earlier studies of this topic, but the term state complexity was not used. For example, in 1963 Lupanov [20] showed that the bound 2 n is tight for the conversion of nondeterministic finite automata (NFA s) to DFA s, and he used the term slozhnost avtomatov, meaning complexity of automata representing the same set of words. The case of languages over a one-letter alphabet was studied in 1964 by Lyubich [21]. Lupanov s result is almost unknown in the English-language literature, and is often attributed to the 1971 paper by Moore [23]. In 1970, Maslov [22] studied the complexity of basic operations on regular languages, and stated without proof some tight bounds for these operations. In the introduction to his paper he states: An important characteristic of the complexity of these sets [of words] is the number of states of the minimal representing automaton. 2 In 1981 Leiss [19] referred to (deterministic) complexity of languages. Some additional references to early works related to this topic can be found in [11, 31], for example. A language is a subset of the free monoid Σ generated by a finite alphabet Σ. If state complexity is a property of a language, then why is it defined in terms of a completely different object, namely an automaton? Admittedly, regular languages and finite automata are closely related, but there is a more natural way to define this complexity of languages, as is shown below. The left quotient, or simply quotient of a language L by a word w is defined as the language w 1 L = {x Σ wx L}. The quotient complexity of L is the number of distinct languages that are quotients of L, and will be denoted by κ(l) (kappa for both kwotient and komplexity). Quotient complexity is defined for any language, and so may be finite or infinite. Since languages are sets, it is natural to define set operations on them. The following are typical set operations: complement (L = Σ \ L), union (K L), intersection (K L), difference (K \ L), and symmetric difference (K L). A general boolean operation with two arguments is denoted by K L. Since languages are also subsets of a monoid, it is also natural to define product, usually called (con)catenation, (K L = {w Σ w = uv,u K,v L}), star (K = i 0 Ki ), and positive closure (K + = i 1 Ki ). The operations union, product and star are called rational or regular. Rational (or regular) languages over Σ are those languages that can be obtained from the set {, {ε}} {{a} a Σ} of basic languages, where ε is the empty word, (or, equivalently, from another basis, such as the finite languages over Σ) using 1 An error in [1] was corrected in [32]. 2 The emphasis is mine.

3 Quotient Complexity of Regular Languages 3 a finite number of rational operations. Since it is cumbersome to describe regular languages as sets for example, one has to write L = ({ε} {a}) {b} one normally switches to regular (or rational) expressions. These are the terms of the free algebra over the set Σ {,ε} with function symbols 3,, and [25]. For the example above, one writes E = (ε a) b. The mapping L from this free algebra onto the algebra of regular languages is defined inductively as follows: L( ) =, L(ε) = {ε}, L(a) = {a}, L(E F) = L(E) L(F), L(E F) = L(E) L(F), L(E ) = (L(E)), where E and F are regular expressions. The product symbol is usually dropped, and languages are denoted by expressions without further mention of the mapping L. Since regular languages are closed under complementation, complementation is treated here as a regular operator. Because regular languages are defined by regular expressions, it is natural to use regular expressions also to represent their quotients; these expressions are their derivatives [4]. First, the ε-function of a regular expression L, denoted by L ε, is defined as follows: a ε = {, if a =, or a Σ; ε, if a = ε. (1) (L) ε = {, if L ε = ε; ε, if L ε =. (2) (K L) ε = K ε L ε, (KL) ε = K ε L ε, (L ) ε = ε. (3) One verifies that L(L ε ) = {ε} if ε L, and L(L ε ) =, otherwise. The derivative by a letter a Σ of a regular expression L is denoted by L a and defined by structural induction: b a = {, if b {,ε}, or b Σ and b a; ε, if b = a. (4) (L) a = L a, (K L) a = K a L a, (KL) a = K a L K ε L a, (L ) a = L a L. (5) The derivative by a word w Σ of a regular expression L is denoted by L w and defined by induction on the length of w: L ε = L, L w = L a, if w = a Σ, L wa = (L w ) a. (6) A derivative L w is accepting if ε L w ; otherwise it is rejecting. One can verify by structural induction that L(L a ) = a 1 L, for all a Σ, and then by induction on the length of w that, for all w Σ, 3 The symbol + is used instead of in [25]. L(L w ) = w 1 L. (7)

4 4 J. Brzozowski Thus every derivative represents a unique quotient of L, but there may be many derivatives representing the same quotient. Two regular expressions are similar [3, 4] if one can be obtained from the other using the following rules: L L = L, K L = L K, K (L M) = (K L) M, (8) L = L, L = L =, εl = Lε = L. (9) Upper bounds on the number of dissimilar derivatives, and hence on the quotient complexity, were derived in [3, 4]: If m and n are the quotient complexities of K and L, respectively, then κ(l) = κ(l), κ(k L) mn, κ(kl) m2 n, κ(l ) 2 n 1. (10) This immediately implies that the number of derivatives, and hence the number of quotients, of a regular language is finite. It seems that the upper bounds in Equation (10), derived in 1962 [3, 4], were the first state complexity bounds to be found for the regular operations. Since the aim at that time was simply to show that the number of quotients of a regular language is finite, the tightness of the bounds was not considered. Of course, the concepts above are related to the more commonly used ideas. A deterministic finite automaton, or simply automaton, is a tuple A = (Q,Σ,δ,q 0,F), where Q is a finite, non-empty set of states, Σ is a finite, non-empty alphabet, δ : Q Σ Q is the transition function, q 0 Q is the initial state, and F Q is the set of final states. The transition function is extended to δ : Q Σ Q as usual. A word w is recognized (or accepted) by automaton A if δ(q 0,w) F. It was proved by Nerode [24] that a language L is recognizable by a finite automaton if and only if L has a finite number of quotients. The quotient automaton of a regular language L is A = (Q,Σ,δ,q 0,F), where Q = {w 1 L w Σ }, δ(w 1 L,a) = (wa) 1 L, q 0 = ε 1 L = L, and F = {w 1 L ε w 1 L}. It should now be clear that the state complexity of a regular language L is the number of states in its quotient automaton, i.e., the number κ(l) of its quotients. This terminology change may seem trivial, but has some nontrivial consequences. For convenience, derivative notation will be used to represent quotients, in the same way as regular expressions are used to represent regular languages. By convention, L ε w always means (L w) ε. 3 Derivation of bounds using quotients Since languages over one-letter alphabets have very special properties, we usually assume that the alphabet has at least two letters. The complexity of operations on unary languages has been studied in [26, 31].

5 Quotient Complexity of Regular Languages 5 In the literature on state complexity, it is assumed that automata A and B accepting languages K and L, respectively, are given. An assumption has to be made that the automata are complete, i.e., that for each q Q and a Σ, δ(q,a) is defined [33]. In particular, if a dead or sink state, which accepts no words, is present, one has to check that only one such state is included [7]. Also, every state must be useful in the sense that it appears on some accepting path [8]. Suppose that a bound on the state complexity of f(k,l) is to be computed, where f is some regular operation. In some cases a DFA accepting f(k,l) is constructed directly, (e.g., Theorems 2.3 and 3.1 in [33]), or an NFA with multiple initial states is used, and then converted to a DFA by the subset construction (e.g., Theorem 4.1 in [33]). Sometimes an NFA with empty-word transitions is used and then converted to a DFA [29]. The constructed automata then have to be proved minimal. Much of this is unnecessary. If quotients are used, the problem of completeness does not arise, since all the quotients of a language are included. A quotient is either empty or useful. If the empty quotient is present, then it appears only once. Since quotients are distinct languages, the set of quotients of a language is always minimal. To find an upper bound on the state complexity, instead of constructing an automaton for f(k,l), we need only find a regular expression for the typical quotient, and then do some counting. This is illustrated below for the basic regular operations. 3.1 Bounds for basic operations The following are some useful formulas for the derivatives of regular expressions: Theorem 1. If K and L are regular expressions, then (L) w = L w, (11) (K L) w = K w L w, (12) (KL) w = K w L K ε L w K ε ul v. (13) w=uv u,v Σ + For the Kleene star, (L ) ε = ε LL, and for w Σ +, (L ) w = (L ) ε ul v L. (14) w=uv u,v Σ Proof : Consider first the boolean operations. Since (K L) w = K w L w, and (L) w = L w, it follows that (K L) w = K w L w.

6 6 J. Brzozowski Equation 13 is easily verified by induction on the length w of a word w Σ. Thus (KL) w consists of K w L and a union (possibly empty) of derivatives of L. When w is in K, then ε K w and L is added to the union. For the star, the claim is obvious when w = ε. For w ε, we first prove that (L ) w = L w (L ) ε u L v L (15) w=uv u,v Σ + by induction on w. Let M = L. For w = a Σ, we have M a = L a L = L a M, by definition. This agrees with (15), because there is no decomposition a = uv with u,v Σ +. Now assume that (15) holds for w, and consider wa: M wa = L wa L ε wl a = L wa L ε w MuL ε va MuL ε ε vl a M w=uv u,v Σ + MuL ε ε v L a MuL ε va w=uv u,v Σ + w=uv u,v Σ + From (15), for w ε, Mw ε = L ε w w=uv Mε ul ε v; thus we have u,v Σ + M wa = L wa Mw ε L a Mu ε L va M = L wa w=uv u,v Σ + M. ε Mx L y wa=xy x,y Σ + M. So the induction step goes through, and we have (15). Note that M = MM; thus w M implies M w = (L ) w L LL = LM, and we have proved a useful alternate version of (15): M w = L w MwL ε M ε ul v M = S(w)M, (16) w=uv u,v Σ + ( where S(w) = L w MwL ε ) w=uv Mε ul v. Finally, note that L w = MεL ε w ; ( u,v Σ ) + hence S(w) = w=uv Mε u,v Σ u L v and we have (14). Theorem 1 can be applied to obtain upper bounds on the complexity of operations. In Theorem 2 below, the second part is a slight generalization of the bound in Theorem 4.3 of [33]. The third and fourth parts are reformulations of the bounds in Theorem 2.3 and 2.4, and of Theorem 3.1 of [33]: Theorem 2. For any languages K and L with κ(k) = m and κ(l) = n, (I) κ(l) = n. (II) κ(k L) mn.

7 Quotient Complexity of Regular Languages 7 (III) Suppose K has k accepting quotients and L has l accepting quotients. (A) If k = 0 or l = 0, then κ(kl) = 1. (B) If k,l > 0 and n = 1, then κ(kl) m (k 1). (C) If k,l > 0 and n > 1, then κ(kl) m2 n k2 n 1. (IV)(A) If n = 1, then κ(l ) 2. (B) If n > 1 and L ε is the only accepting quotient of L, then κ(l ) = n. (C) If n > 1 and L has l > 0 accepting quotients not equal to L, then κ(l ) 2 n n l 1. Proof : The first part is well-known, and the second follows from (12). For the product, if k = 0 or l = 0, then KL = and κ(kl) = 1. Thus assume that k,l > 0. If n = 1, then L = Σ and w K implies (KL) w = Σ. Thus all k accepting quotients of K produce the one quotient Σ in KL. For each rejecting quotient of K, we have two choices for the union of quotients of L in (13): the empty union or Σ. If we choose the empty union, we can have at most m k quotients of KL. Choosing Σ results in (KL) w = Σ, which has been counted already. Altogether, there are at most 1 + m k quotients of KL. Suppose now that k,l > 0 and n > 1. If w / K, then we can choose K w in m k ways, and the union of quotients of L in 2 n ways. If w K, then we can choose K w in k ways, and the set of quotients of L in 2 n 1 ways, since L is then always present. Thus we have (m k)2 n + k2 n 1. For the star, if n = 1, then L = or L = Σ. In the first case, L = ε, and κ(l ) = 2; in the second case, L = Σ and κ(l ) = 1. Now suppose that n > 1; hence L has at least one accepting quotient. If L is the only accepting quotient of L, then L = L and κ(l ) = κ(l). Now assume that n > 1 and l > 0. From (14), every quotient of L by a nonempty word is a union of a subset of quotients of L, followed by L. Moreover, that union is non-empty, because (L ) ε ε L w is always present. We have two cases: (I) Suppose L is rejecting. Then L has l accepting quotients. (A) If no accepting quotient of L is included in the subset, then there are 2 n l 1 such subsets possible, the union being non-empty because L w is always included. (B) If an accepting quotient of L is included, then ε (L ) w, (L ) ε w = ε, and L = (L ) ε wl ε is also included. We have 2 l 1 non-empty subsets of accepting quotients of L and 2 n l 1 subsets of rejecting quotients, since L is not counted. Adding 1 for (L ) ε, we have a total of 2 n l 1 + (2 l 1)2 n l = 2 n n l 1. (II) Suppose L is accepting. Then L has l + 1 accepting quotients. (A) If there is no accepting quotient, there are 2 n l 1 1 non-empty subsets of rejecting quotients. (B) If an accepting quotient of L is included, then L is included, and 2 n 1 subsets can be added to L.

8 8 J. Brzozowski We need not add (L ) ε, since ǫ LL = LL in this case, and this has already been counted. The total is 2 n n l 1. The worst-case bound of 2 n n l 1 occurs in the first case only. 3.2 Witnesses to bounds for basic operations Finding witness languages showing that a bound is tight is often challenging. However, once a guess is made, the verification can be done using quotients. Let w a be the number of a s in w, for a Σ and w Σ. Unary, binary, and ternary languages are languages over a one-, two-, and three-letter alphabet, respectively. Union and Intersection If we have a bound for intersection, then for union we can use the fact that κ(k L) = κ(k L) = κ(k L); thus the pair (K,L) is a witness for union. Similarly, given a witness for union, we also have a witness for intersection. The upper bound mn for the complexity of intersection was observed in by Rabin and Scott [27]. Binary languages K = {w {a,b} w a m 1 mod m} and L = {w {a,b} w b n 1 mod n} have quotient complexities m and n, respectively. In 1970 Maslov [22] stated without proof that K L meets this upper bound mn. Yu, Zhuang and K. Salomaa [33], used similar languages K = {w {a,b} w a 0 mod m} and L = {w {a,b} w b 0 mod n} for intersection, apparently unaware of [22]. Hricko, Jirásková and Szabari [11] showed that a complete hierarchy of quotient complexities of binary languages exists between the minimum complexity 1 and the maximum complexity mn. More specifically, it was proved that for any integers m,n,α such that m 2, n 2 and 1 α mn, there exist binary 5 languages K and L such that κ(k) = m, κ(l) = n, and κ(k L) = α, and the same holds for intersection. For a one-letter alphabet Σ = {a}, Yu showed that the bound can be reached if m and n are relatively prime [31]. The witnesses are K = (a m ) and L = (a n ). For other cases, see the paper by Pighizzini and Shallit [26]. Set difference For set difference we have κ(k \ L ) = κ(k L ); thus the pair (K,L ) is a witness. Symmetric difference For symmetric difference, let m, n 1, let K = (b a) m 1 (a b) and let L = (a b) n 1 (a b). There are mn words of the form a i b j, where 0 i m 1 and 0 j n 1. We claim that all the quotients of K L by these words are distinct. Let x = a i b j and y = a k b l. If i < k, let u = a m 1 k b n. Then xu / K, yu K, and xu,yu L, 4 The work was done in 1957, but published in The proof in [11] is for ternary languages; a proof for the binary case can be found in [10].

9 Quotient Complexity of Regular Languages 9 showing that xu K L, and yu / K L, i.e., that (K L) x (K L) y. Similarly, if j < l, let v = a m b n 1 l. Then xv K L, but yv / K L. Therefore all the quotients of K L by these mn words are distinct. For a one-letter alphabet, the witnesses are K and L as in the case of union above. Other boolean functions There are six more two-variable boolean functions that depend on both variables: K L = K L, K L = K L, K L = K \ L, K L = L\K, K L = L \ K, and K L. The witnesses for these functions can be found using the four functions above. Product The upper bound of m2 n 2 n 1 was given by Maslov in 1970 [22], and he stated without proof that it is tight for binary languages K = {w {a,b} w a m 1 mod m} and L = (a b) n 2 (a b)(b a(a b)). The bound was refined by Yu, Zhuang and K. Salomaa [33] to m2 n k2 n 1, where k is the number of accepting quotients of K. Jirásek, Jirásková and Szabari [12] proved that, for any integers m,n,k such that m 2, n 2 and 0 < k < m, there exist binary languages K and L such that κ(k) = m, κ(l) = n, and κ(kl) = m2 n k2 n 1. Furthermore, Jirásková [15] proved that, for all m, n, and α such that either n = 1 and 1 α m, or n 2 and 1 α m2 n 2 n 1, there exist languages K and L with κ(k) = m and κ(l) = n, defined over a growing alphabet, such that κ(kl) = α. For a one-letter alphabet, mn is a tight bound for product if m and n are relatively prime [33]. The witnesses are K = (a m ) a m 1 and L = (a n ) a n 1. See also [26]. Star Maslov [22] stated 6 without proof that κ(l ) 2 n 1 +2 n 2, and provided a binary language meeting this bound. Three cases were considered by Yu, Zhuang and K. Salomaa [33]: n = 1. If L =, then κ(l) = 1 and κ(l ) = 2. If L = Σ, then κ(l ) = 1. n = 2. L = {w {a,b} w a 1 mod 2} has κ(l) = 2, and κ(l ) = 3. n > 2. Let Σ = {a,b}. Then L = (b aσ n 1 ) aσ n 2 has n quotients, one of which is accepting, and κ(l ) = 2 n n 2. This example is different from Maslov s. Moreover, Jirásková [13] proved that, for all integers n and α with either 1 = n α 2, or n 2 and 1 α 2 n 1 +2 n 2, there exists a language L over a 2 n letter alphabet such that has κ(l) = n and κ(l ) = α. For a one-letter alphabet, n 2 2n + 2 is a tight bound for star [33]. The witness is L = (a n ) a n 1. See also [26]. 6 The bound is incorrectly stated as 2 n n 2 1, but the example is correct.

10 10 J. Brzozowski 4 Generalization of non-returning state A quotient L w of a language L is uniquely reachable if L x = L w implies that x = w. If L wa is uniquely reachable for a Σ, then so is L w. Thus, if L has a uniquely reachable quotient, then L itself is uniquely reachable by the empty word, i.e., the minimal automaton of L is non-returning 7. Thus the set of uniquely reachable quotients of L is a tree with root L, if it is non-empty. We now apply the concept of uniquely reachable quotients to boolean operations and product. Theorem 3. Suppose κ(k) = m, κ(l) = n, K and L have m u and n u uniquely reachable quotients, respectively, and there are r words w i such that both K wi and L wi are uniquely reachable. If is a boolean operator, then κ(k L) mn (α + β + γ), where (17) α = r(m+n) r(r+1); β = (m u r)(n (r+1)); γ = (n u r)(m m u 1). (18) If K has k accepting quotients, t of which are uniquely reachable, and s rejecting uniquely reachable quotients, then κ(kl) m2 n k2 n 1 s(2 n 1) t(2 n 1 1). (19) Proof : Suppose the quotients of K and L are K 1,...,K m and L 1,...,L n, respectively. Without loss of generality, we can assume that this numbering is such that K wi = K i and L wi = L i, for each i = 1,...,r. Moreover, assume that the remaining m u r uniquely reachable quotients of K are numbered K r+1,...,k mu, and the remaining n u r uniquely reachable quotients of L are L mu+1,...,l mu+(n u r). Because of (12), the number of quotients of K L is bounded from above by the number of pairs (K w,l w ). Consider first K w1 = K 1 and L w1 = L 1. Since K 1 can appear only with L 1 in any pair, we know that the n 1 pairs of the form (K 1,L j ), j > 1, and the m 1 pairs of the form (K i,l 1 ), i > 1 will never appear. Next, if K 2 and L 2 are both uniquely reachable, then we know that the n 2 pairs (K 2,L j ), j > 2, and m 2 pairs (K i,l 1 ), i > 2 will not appear. Finally, the n r pairs (K r,l j ), j > r, and the m r pairs (K i,l r ), i > r, will not appear. Thus, we have the following reduction due to the r pairs: α = m m r + n n r = r(m + n) r(r + 1). We have now examined r of the m u quotients that are uniquely reachable in K. For each of the remaining m u r uniquely reachable quotients of K we can eliminate n (r+1) pairs of the form (K i,l j ), r+1 < i < m u, where K w = K i, but L w L j. This yields the second part of the reduction: β = (m u r)(n (r+1)). For each of the n u r uniquely reachable quotients of L, we can eliminate m m u 1 pairs of the form (K i,l j ), r + 1 < i < m u, where L w = L j, but K w K i, obtaining the third reduction: γ = (n u r)(m m u 1). 7 The term non-returning suggests that once a state is left it cannot be visited again. However, such non-returning states are not necessarily uniquely reachable.

11 Quotient Complexity of Regular Languages 11 For the product, each quotient of KL corresponds to one of the m quotients of K together with a subset of the n quotients of L. If K w is rejecting, then it can appear with 2 n subsets of quotients of L. This gives (m k)2 n such possible quotients of KL. If a rejecting quotient of K is uniquely reachable, then it can appear with only one subset. Hence there is a savings of s(2 n 1). If K w is accepting, then L is always present in (KL) w. Hence there are at most k2 n 1 such possible quotients of KL. But, if t of the accepting quotients of K are uniquely reachable, then each can appear with only one subset of quotients of L that does not include L, for a saving of t(2 n 1 1). Altogether, we have at most (m k)2 n s(2 n 1) + k2 n 1 t(2 n 1 1) quotients of KL, as claimed. The following observation was stated for union and intersection of finite languages in [31]; we add the suffix-free case: Corollary 4. If K and L are non-empty and finite or suffix-free and κ(k) = m > 1, κ(l) = n > 1, then κ(k L) mn (m + n 2). Proof : The quotients K ε and L ε are uniquely reachable if K and L are finite. This also holds if K and L are suffix-free. For suppose L ε = L w. Since L is non-empty, we have ε L x for some x Σ. Then also ε L wx and both wx and x are in L, contradicting suffix-freeness. The bound mn (m + n 2) for union of suffix-free languages was shown to be tight for quinary languages by Han and Salomaa [7]. It is also tight for the binary languages K = a((ba ) m 3 b) (ba ) m 3 and L = a((a b) n 3 b) (a b) n 3, as shown recently by Jirásková and Olejár [17]. b b 1 a b 3 2 a a, b 4 7 a, b a 3 b 5 1 a b a 6 b a, b 2 b a a b a 4 a, b 5 (a) (b) Figure 1. Illustrating unique reachability. Example 5. The automaton of Fig. 1 (a) accepting K has m = 7 and four uniquely reachable states: 1, 2, 3, and 4. The automaton of Fig. 1 (b) accepting L has n = 5 and three uniquely reachable states: 1, 2, and 5. In pairs (1,1) and (2,2) both states are reachable by the same word (ε and b, respectively); hence r = 2. The m n = 7 5 table of all pairs is shown below, where uniquely reachable states are in boldface type. We have α = 18, where the removed pairs are all the

12 12 J. Brzozowski pairs in the first two rows and columns, except (1,1) and (2,2). Next, β = 4, and we remove the pairs (3,4), (3,5), (4,3) and (4,5) from rows 3 and 4. Finally, γ = 2, and we remove the pairs (6,5) and (7,5) from column 5. (1,1) (1, 2) (1, 3) (1, 4) (1, 5) (2, 1) (2,2) (2, 3) (2, 4) (2, 5) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (5, 1) (5, 2) (5, 3) (5, 4) (5,5) (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (7, 1) (7, 2) (7, 3) (7, 4) (7, 5) Altogether, we have removed 24 states from K L, leaving 11 possibilities. The minimal automaton of K L has 8 states. Notice that state 7 corresponds to the quotient Σ. Since Σ L w = Σ for all w, we need to account for only one pair (7, x), and we could remove the remaining four pairs. However, we have already removed pair (7,5) by Theorem 3. Hence, there are only three pairs left to remove, and we have an automaton with 8 states. More will be said about the effects of Σ later. It is also possible to use Theorem 3 if K has some uniquely reachable quotients and L has none, or when L is completely unknown. If n u = 0, then r = 0, α = 0, β = m u (n 1), and γ = 0. Then, for any L, κ(k L) mn m u (n 1). (20) For example, for any L with n = 101 and K as in Fig. 1 (a), κ(k L) 307, instead of the general bound 707. Let K and L be the automata of Fig. 1 (a) and (b), respectively. Then the general bound on κ(kl) is 192. Here s = 3 (states 1, 2, and 4), and t = 1 (state 3). By Theorem 3 the bound is reduced by = 108 to 84. The actual quotient complexity of KL is 14. The general bound for LK is 512, the reduced bound is 195, and the actual quotient complexity is Languages with ε, Σ +,, or Σ as quotients In this section we consider the effects of the presence of special quotients in a language. In particular, we study the quotients ε, Σ +,, and Σ. Theorem 6. If κ(k) = m, κ(l) = n, and K and L have k > 0 and l > 0 accepting quotients, respectively, then (I) If K and L have ε as a quotient, then κ(k L) mn 2. κ(k L) mn (2m + 2n 6). κ(k \ L) mn (m + 2n k 3). κ(k L) mn 2.

13 Quotient Complexity of Regular Languages 13 (II) If K and L have Σ + as a quotient, then κ(k L) mn 2. κ(k L) mn (2m + 2n 6). κ(k \ L) mn (2m + l 3). κ(k L) mn 2. (III) If K and L have as a quotient, then κ(k L) mn (m + n 2). κ(k \ L) mn n + 1. (IV) If K and L have Σ as a quotient, then κ(k L) mn (m + n 2). κ(k \ L) mn m + 1. (V) If L has ε as a quotient, then κ(l R ) 2 n Proof : If L has Σ + as a quotient, then κ(l R ) 2 n If L has as a quotient, then κ(l R ) 2 n 1. If L has Σ as a quotient, then κ(l R ) 2 n 1. Moreover, the effect of these quotients on complexity is cumulative. For example, if L R has both and Σ, then κ(l R ) 2 n 2, if L R has both and Σ +, then κ(l R ) 2 n 3 + 1, etc. (I) The quotient of ε by every non-empty word is ; thus, if L has ε, it also has. Since ε ε = ε = ε = ε, we subtract two quotients for union. For intersection, for any K u and L v, K u = L v =. This eliminates m 1+n 1 possibilities. Moreover, if K u,l v are rejecting, then K u ε = ε L v =. This removes another m k 1+n l 1 possibilities. If K u,l v are accepting, then K u ε = ε L v = ε, and k 1+l 1 quotients are removed. Altogether, κ(k L) mn (2m + 2n 6). For set difference, since K has and L w = for all w, this saves n 1 quotients. Since K has ε, if ε L w, then ε L w = ε; otherwise ε L w =. This saves another n 1 quotients. If L has ε and, then L has Σ + and Σ. For each rejecting quotient K w, we have K w Σ = K w Σ +. This saves another (m k) 1 quotients of K \ L. Since ε = ε = ε, and ε ε = = we can subtract two quotients for symmetric difference. (II) The proofs are dual to those of (i): The quotient of Σ + by every non-empty word is Σ ; thus, if L has Σ +, it also has Σ. Since Σ + Σ + = Σ Σ + = Σ + Σ = Σ +, we subtract two quotients for intersection. For union, for any K u and L v, K u Σ = Σ L v = Σ. This eliminates m 1 + n 1 possibilities. Moreover, if K u,l v are accepting, then K u Σ + = Σ + L v = Σ. This removes another k 1 + l 1

14 14 J. Brzozowski possibilities. If K u,l v are rejecting, then K u Σ + = Σ + L v = Σ +, removing m k 1 + n l 1 quotients. For set difference, if L has Σ + and Σ, then L has ε and. Since K w = for all w, this saves m 1 quotients. Also, if ε K w, then K w ε = ε, and otherwise K w ε =. This saves another m 1 quotients. Finally, for each accepting quotient L w, we have Σ L w = Σ + L w. This saves another l 1 quotients. Since Σ Σ + = Σ + Σ = ε and Σ Σ = Σ + Σ + =, we can subtract two quotients for symmetric difference. (III) Since K w = L w =, we can subtract (m 1)+(n 1) quotients. For difference, since L w = for all w, this saves n 1 quotients. (IV) Dual to iii. (V) The subset construction is used to convert the nondeterministic automaton N R, which is the reverse of the quotient automaton A recognizing L, to a deterministic automaton A R recognizing L R. In that construction, the state of N R corresponding to is not reachable from the set of initial states, and the state corresponding to ε appears only in the initial set of states, but is not reachable by any non-empty word from that set. Dually, the state corresponding to Σ appears in all the sets reachable from the set of initial states, and Σ + appears in all the sets reachable from the set of initial states, except the set of initial states. Corollary 7. If K and L are both non-empty and both suffix-free with κ(k) = m and κ(l) = n, then κ(k L) mn 2(m + n 3). Proof : We have shown in Corollary 4 that κ(k L) mn (m + n 2) by removing all pairs (K,L u ) and (K v,l), where L u L and K v K. If K is suffix-free, then it must have as a quotient [7], and the same holds for L. Thus all pairs of the form (K w, ) and (,L w ) are equivalent to (, ). We have already removed (K, ) and (,L) by unique reachability. Hence we can remove a further (m + n 2) 2 quotients, for a total of 2(m + n 3). It is shown in [7] that the bound can be reached with K = {#w w {a,b}, w a 0 mod m 2}, L = {#w w {a,b}, w b 0 mod n 2}. It was recently proved in [17] that this bound can be reached by the binary languages given after Corollary 4. Proposition 8. If κ(l) = n 3, L has l > 0 accepting quotients, and L has ε as a quotient, then κ(l ) 2 n n l

15 Quotient Complexity of Regular Languages 15 Proof : If L has ε, then it also has. From (14), every quotient of L by a non-empty word is a union of a non-empty subset of quotients of L, followed by L. We have two cases: (I) Suppose L is rejecting. (A) If no accepting quotient is included, then there are 2 n l 1 1 nonempty subsets of non-empty rejecting quotients plus the subset consisting of the empty quotient alone, for a total of 2 n l 1. (B) If an accepting quotient is included in the subset, then so is L. We can add the subset {ε} or any non-empty subset S of accepting quotients that does not contain ε, since S {ε} is equivalent to S. Thus we have 2 l 1 subsets of accepting quotients. To this we can add 2 n l 2 rejecting subsets, since the empty quotient and L need not be counted. The total is 2 l 1 2 n l 2 = 2 n 3. Adding 1 for (L ) ε, we have a total of 2 n n l (II) Suppose L is accepting. Since n 3, we have L ε. (A) If there is no accepting quotient, there are 2 n l 1 subsets, as before. (B) If an accepting quotient is included, then L is included and L itself is sufficient to guarantee that (L ) w is accepting. Since L ε = L = L, we also exclude ε and. Thus any one of the 2 n 3 subsets of the remaining quotients can be added to L. The total is 2 n n l 1. We need not add (L ) ε, since it is LL which has been counted already. The worst-case bound of 2 n n l occurs in the first case only. 6 Prefix-free languages As another example of the application of the quotient methods, we now derive the bounds for basic operations on prefix-free languages [8]. Proposition 9. If K and L are both prefix-free and non-empty with κ(k) = m and κ(l) = n, then (I) κ(k L) mn 2, (II) κ(k L) mn 2, (III) κ(k L) mn (2m + 2n 6), (IV) κ(kl) m + n 2, (V) κ(l ) n. Proof : If L is non-empty, then it has at least one accepting quotient L w. If L is also prefix-free, then ε L w implies L w = ε. Thus both K and L have ε as a quotient. Parts (i) (iii) follow directly from Theorem 6. For (iv), consider Equation (13). If K = ε, then m = 2, KL = L and κ(kl) = n. If n = 1, then L can only be Σ, which is not prefix-free. If n 2, then n m + n 2. Assume now that ε K; hence the term K ε L w is missing

16 16 J. Brzozowski in (13). Consider any w ε. If ε K w, K w, and ε K u for some proper prefix u of w = uv, then K cannot be prefix-free, because it contains u and wx, for some x Σ +. Thus, if K w is rejecting and non-empty, then (KL) w = K w L; there are m 2 such quotients, the remaining two quotients being ε and. If K w = ε, then (KL) w = L. Then (KL) wx = L x, for all x Σ, and there are n such quotients. If K w =, then (KL) w =, and this quotient has already been counted in the case where K w = ε and L x =. Thus the total is at most m + n 2. For (v), we have (L ) ε = L, and for w Σ +, consider Equation (14). By the argument we used in (iv), if L w is rejecting and non-empty, then (L ) w = L w L, and there are n 2 such quotients. If L w is accepting, then L w = ε, and (L ) w = L w L = L = (L ) ε. Finally, if L w =, then (L ) w may be empty, if for every prefix u of w, L u is rejecting. Thus the bound is n. It is shown in [8] that the bounds are tight: Let K = (a m 2 ) b and L = (a n 2 ) b. Then κ(kl) = m + n 2, and κ(l ) = n. If Σ = {a,b,c} and K = {wc w {a,b} and w a 0 mod m}, L = {wc w {a,b} and w b 0 mod n}, then κ(k L) = mn (2m + 2n 6). Finally, if Σ = {a,b,c,d} and then κ(k L) = mn 2. K = {wc w {a,b,d} and w a 0 mod m}, L = {wd w {a,b,c} and w b 0 mod n}, 7 Conclusions Quotients provide a uniform approach for finding upper bounds for the complexity of operations on regular languages, and for verifying that particular languages meet these bounds. It is hoped that this is a step towards a theory of complexity of languages and automata. Acknowledgments I am very grateful to Galina Jirásková for correcting several errors in early versions of this paper, suggesting better examples, improving proofs, and helping me with references, in particular, with the early work on complexity. I thank Sheng Yu for his help with references, and for answering many of my questions on complexity. I also thank Baiyu Li, Shengying Pan, and Jeff Shallit for their careful reading of the manuscript. References [1] J. C. Birget. Intersection of regular languages and state complexity. ACM SIGACT News, 22(2), 49, 1991.

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